Multiple Random Variables Problems

A random point $(X,Y)$ is distributed uniformly on the square with vertices $(1, 1),(1,-1),(-1,1),$ and $(-1,-1)$. That is, the joint pdf is $f(x,y)=\frac{1}{4}$ on the square. Determine the probabilities of the following events.

$X^2 + Y^2 < 1$

$2X-Y>0$

$|X+Y|<1$ (modified since the original $|X+Y|<2$ is trivial.)

Solutions:

$X^2 + Y^2 < 1$
We need to consider the boundary of this inequality first in the unit square, so below is the plot of $X^2 + Y^2 = 1$,

Hence, we are interested in the area of the ellipse above since $X^2 + Y^2$ is less than 1. To compute the area of the ellipse, notice that the regions in the 4 quadrants are identical except on the orientation, thus we can compute the area of the first quadrant then we simply multiply this by 4 to cover the overall area of the said geometric.
\begin{equation}\nonumber
\begin{aligned}
P(X^2 + Y^2 < 1) &= 4\int_{0}^{1}\int_{0}^{\sqrt{1 - x^2}}\frac{1}{4}\operatorname{d}y\operatorname{d}x\\
&= \int_{0}^{1}y\Bigg|_{y=0}^{y=\sqrt{1 - x^2}}\operatorname{d}x = \int_{0}^{1}\sqrt{1 - x^2}\operatorname{d}x\\
&=\left(\frac{x}{2} \sqrt{- x^{2} + 1} + \frac{1}{2} \operatorname{sin}^{-1}{\left (x \right )}\right)\Bigg|_{x=0}^{x=1}\\
&=\frac{\pi}{4}-0=\frac{\pi}{4}.
\end{aligned}
\end{equation}
Confirm this using python symbolic computation,

Given $2X-Y>0$, we have
\begin{equation}\nonumber
P(2X-Y>0)=P(-Y>-2x) = P(Y<2X)
\end{equation}
The plot of $Y=2X$ is shown below,

Find the value of $C$.
To solve for the value of $C$ we integrate the given pdf first for $x$ and $y$, that is
\begin{equation}\nonumber
\begin{aligned}
1&=\int_{0}^{1}\int_{0}^{2}C (x+2y)\operatorname{d}x\operatorname{d}y=
C\int_{0}^{1} \left(\frac{x^2}{2}+2xy\right)\Bigg|_{x=0}^{x=2}\operatorname{d}y\\
&=C\int_{0}^{1}(2+4y)\operatorname{d}y=C\left(2y+4\frac{y^2}{2}\right)\Bigg|_{y=0}^{y=1}\\
1&=4C\Rightarrow C=\frac{1}{4}
\end{aligned}
\end{equation}